Complex numbers

  • #1

Homework Statement



http://img139.imageshack.us/img139/999/ksknl6.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

Could someone confirm that the a) part is correct, and if it is, then what is the next step? :)
And im not exactly sure what to do in the b) part.

I would really appriciate if someone could help me out with this. :)

Thanks in advance.
 
Last edited by a moderator:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
NO, (a) is not correct because you haven't answered the question! I think you have misunderstood what is asked. If you were asked to find [itex]^4\sqrt{16}[/itex], you wouldn't answer "161/4"!

Don't change from polar to rectangular form until after you have done the root! One important reason for using polar form for complex numbers is the fact that
[tex][r (cos(\theta)+ i sin(\theta)]^n= r^n(cos(n\theta)+ i sin(n\theta))[/tex]

For n= 1/4,
[tex](7(cos(\pi/2)+ i sin(\pi/2))^{1/4}= 7^{1/4}(cos(\pi/8)+ i sin(\pi/8))[/tex]
Also, since adding [itex]2\pi[/itex] to the argument doesn't change the complex number, and [itex]2\pi/4= \pi/2[/itex] another fourth root is
[tex]7^{1/4}(cos(\pi/4+ \pi/2)+ i sin(\pi/4+ \pi/2))[/tex]
yet another is
[tex]7^{1/4}(cos(\pi/4+ \pi)+ i sin(\pi/4+ \pi))[/tex]
and, finally,
[tex]7^{1/4}(coS(\pi/4+ 3\pi/2)+ i sin(\pi/4+ 3\pi/2))[/tex]

Those are the four fourth roots of [itex]7(cos(\pi/2)+ i sin(\pi/2))= 7i[/itex]
 
  • #3
Nissen, Søren Rune
Edit: Shows what I know. Listen to HallsofIvy instead
 
Last edited by a moderator:

Related Threads on Complex numbers

  • Last Post
Replies
4
Views
898
  • Last Post
Replies
2
Views
852
M
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
11
Views
1K
  • Last Post
Replies
8
Views
2K
Top